Question

Nickel has a face-centered cubic unit cell. The density of nickel is 6.84 \(\mathrm{g} / \mathrm{cm}^{3} .\) Calculate a value for the atomic radius of nickel.

Short answer

The atomic radius of nickel is approximately \(1.24 \times 10^{-8}\) cm.

Step-by-step solution

List the given information

Density of nickel (ρ) = 6.84 g/cm³ Unit cell type: face-centered cubic (fcc)

Calculate the number of atoms in a face-centered cubic unit cell

In an fcc unit cell, each corner atom is shared by 8 adjacent unit cells, and each face atom is shared by 2 adjacent unit cells. Therefore, the number of atoms in an fcc unit cell can be calculated as: Number of atoms = (Number of corner atoms × 1/8) + (Number of face atoms × 1/2) Number of atoms = (8 × 1/8) + (6 × 1/2) Number of atoms = 4

Write down the formula for the density of the face-centered cubic unit cell

The formula for the density of an fcc unit cell is: \(ρ = \frac{4 \times M}{a^{3} \times N_{A}}\) Where: ρ = Density M = Molar mass of nickel (g/mol) a = Edge length of the unit cell (cm) Nₐ = Avogadro's number (6.022 x 10²³ atoms/mol) We need to find the edge length (a) and use it to determine the atomic radius.

Find the molar mass of nickel

The molar mass of nickel (Ni) is 58.69 g/mol.

Rearrange the formula to solve for the edge length (a) of the unit cell

We can rearrange the density formula to solve for a: \(a^{3} = \frac{4 \times M}{ρ \times N_{A}}\) Next, we plug in the values and solve for the edge length: \(a^{3} = \frac{4 \times 58.69 \text{ g/mol}}{6.84 \text{ g/cm}^{3} \times 6.022 \times 10^{23} \text{ atoms/mol}}\) \(a^{3} = \frac{233.76 \text{ g/mol}}{6.84 \text{ g/cm}^{3} \times 6.022 \times 10^{23} \text{ atoms/mol}}\) \(a^{3} \approx 5.94 \times 10^{-23} \text{ cm}^{3}\) \(a \approx \sqrt[3]{5.94 \times 10^{-23} \text{ cm}^{3}}\) \(a \approx 3.93 \times 10^{-8} \text{ cm}\)

Calculate the atomic radius using the edge length

In an fcc unit cell, the relationship between edge length (a) and atomic radius (r) is: \(a = 2 \sqrt{2}r\) Now, we can solve for the atomic radius: \(r = \frac{a}{2 \sqrt{2}}\) \(r = \frac{3.93 \times 10^{-8} \text{ cm}}{2 \sqrt{2}}\) \(r \approx 1.24 \times 10^{-8} \text{ cm}\) So, the atomic radius of nickel is approximately \(1.24 \times 10^{-8}\) cm.

Key concepts

Atomic Radius

When discussing the atomic structure, the atomic radius is an important parameter. The atomic radius is a measure of the size of an atom. It represents the distance from the center of the nucleus to the outermost shell of electrons. In the case of nickel, which has a face-centered cubic (fcc) unit arrangement, determining this size can be particularly interesting.
To find the atomic radius of nickel from its crystal structure, we need to use the edge length of the unit cell. In an fcc unit cell, which nickel exhibits, there is a specific relationship:

• The edge length of the unit cell (\(a\)) relates to the atomic radius (\(r\)) through the equation \(a = 2 \sqrt{2} r\).

By rearranging this equation, you can solve for the atomic radius: \(r = \frac{a}{2 \sqrt{2}}\). Using this formula allows us to precisely calculate the atomic radius given the edge length of the cell. This is vital for further understanding many of nickel's physical properties.

Density Calculation

The density of a substance is a fundamental concept that refers to how much mass is contained in a given volume. For nickel's crystal structure, you may use its density to find other structural details such as the atomic radius. Nickel's density is given as 6.84 \(\text{g/cm}^3\).
For a face-centered cubic (fcc) unit cell, the density can be calculated using the formula: \(\rho = \frac{4 \times M}{a^{3} \times N_{A}}\)where:

• \(\rho\) is the density,
• \(M\) is the molar mass,
• \(a\) is the edge length of the unit cell,
• \(N_{A}\) is Avogadro's number.

By rearranging this formula, we can solve for the edge length \(a\), which in turn is used to derive the atomic radius. This step is key in bridging the relationship between supplied physical properties and the inherent structural characteristics of nickel.

Avogadro's Number

Avogadro's number is a crucial constant in chemistry and materials science. Represented as 6.022 x 10²³ atoms/mol, it corresponds to the number of atoms or molecules in one mole of a substance. This value is essential for calculations involving atomic scale materials.
In the context of the face-centered cubic unit cell for nickel, Avogadro's number helps bridge the gap between atomic and macroscopic scales. When calculating the density of the fcc structure, Avogadro's number allows us to relate the molar mass of an element to the number of atoms in the crystal structure.
This enables precise calculations, such as determining how many individual atoms contribute to the overall properties of a material, thereby revealing insights into the atomic scale features of substances like nickel.

Molar Mass of Nickel

Molar mass is the mass of one mole of a given substance and is expressed in g/mol. For nickel, this is 58.69 g/mol. The molar mass is intrinsic to calculating various properties such as density of a crystalline structure.
In the fcc unit cell for nickel, the molar mass helps in calculating the density, as seen in the formula: \(\rho = \frac{4 \times M}{a^{3} \times N_{A}}\)where \(M\) is the molar mass.
The molar mass is crucial in these calculations because it provides a direct link between the microscopic world of atoms and the macroscopic properties engineers and scientists can measure. It helps quantify how much nickel, in mass, is packed into a specific volume in its crystalline state. Understanding this concept is vital, especially when you study materials science and solid-state physics, where precise calculations are necessary for material design and application.